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Name Table 18.2 (continued) 18.20 Solutions Tested 0,10 M HC HO, pH (H (M) [C,H,O, 1(M) K I Percent Dissociation 12.92 (1.200103 120x103 0,010 M HCH B34 535 | 67 + Question 3.54 4.5740 67410" 2200S 4.57% 3.74 (181810-4 181410" Novio's 18,10% Write the net ionic equation describing the dissociation of acetic acid in aqueous solution. н»,0 , 50, +1° Question Calculate (H') and (C,H,O," for each concentration above, showing a sample cal- culation here. Question Write the equilibrium constant expression for acetic acid, and calculate a Kvalue and an apparent percent dissociation for each concentration, showing sample calculations here. Question What is the effect of dilution on the percent dissociation? Why should this behavior be expected? Instructor's Initials:

Answer 1

+ Explanation: Concentration of H* and C,H,O, From the definition of pH: pH = -log[H+] We can find the concentration of H' from the measured pH: [H+] = 10 pH For the first solution, for example, the measured pH is 2.92 and [H') is: [H+] = 10-2.92 = 1.20 x 10-3 M From the reaction, we see that the concentration of H" is essentially the same as that of C2H20, at equilibrium. Hence, the concentration of C2H20, for this solution is also 1.20 x 10°M. +

Equilibrium constants: The equilibrium constant for this reaction is: K - [H+][C2H30,] Ke=[HC2H3021 The initial quantity of HC2H20, dissolves into x moles of H and x moles of C2H20, as seen in the reaction. Hence, at equilibrium, [H+] = x, [C2H303] = x and [HC2H302] = x; – X, where Xi is the initial concentration of acetic acid. Thus, we can write the equilibrium constant as: Ke = We know X from the calculations we made in the previous part. For example, for the second solution tested: X; = 0.010 M and x = [H+] = 4.57 x 10-4 M: K = (4.57x10-4)(4.57x10-4) . ne = 0.010-457x10-4 = 2.2 * 10%

[HC2H302] Percent dissociation: The percent dissociation is: %dis = 100 CM For the first solution, [C2H302] = 1.20 x 10-3 M and [HC2H302]; = 0.10 M: %dis = 100 1.20-10-3 = 1.2% 0.10 Effect of dilution on dissociation: As we can see, at higher dilution rates (that is, lower initial concentrations of HC,H2O2), the percent dissociation increases. This should be expected because molecules require energy to break bonds and dissociate and they take this energy from water molecules. If there are less water molecules per

molecule of acetic acid, more molecules of acetic acid will remained un- dissociated due to the lack of energy to dissociate.